Question

a circuit includes a capacitor that charges over time. if the resistance of the circuit is 3.0 ω, the capacitance is 2.2 f, and the potential of the circuit is 5.0 v, what is the potential over the capacitor after a time of 5.0 seconds, assuming the capacitor is initially uncharged? 0 4.78 v 0 0.0 v 0 2.7 v 0 5.0 v

Explanation:

Step1: Recall RC - circuit charging formula

The formula for the potential difference across a capacitor in an RC - circuit during charging is $V_C(t)=V_0(1 - e^{-\frac{t}{RC}})$, where $V_0$ is the emf of the battery, $t$ is the time, $R$ is the resistance, and $C$ is the capacitance.

Step2: Identify given values

We are given that $V_0 = 5.0\ V$, $t = 5.0\ s$, $R=3.0\ \Omega$, and $C = 2.2\ F$. First, calculate the time - constant $\tau=RC$.
$\tau=RC=3.0\times2.2 = 6.6\ s$.

Step3: Substitute values into the formula

$V_C(t)=V_0(1 - e^{-\frac{t}{RC}})=5.0\times(1 - e^{-\frac{5.0}{6.6}})$.
Calculate $e^{-\frac{5.0}{6.6}}=e^{- 0.7576}\approx0.469$.
Then $V_C(t)=5.0\times(1 - 0.469)=5.0\times0.531 = 2.655\approx2.7\ V$.

Answer:

2.7 V